How to finish proving this sequence converges?

Question

Here's what I'm tasked with showing:

Let $(a_n)$ be a convergent sequence with $a_n\rightarrow a$ as $n\rightarrow\infty$. By the Algebraic Limit Theorem, we know that $(a_n^2)\rightarrow a^2$. Now prove this using the definition of convergence.

In doing so, I have the following:

Let $\epsilon>0$ be given. Choose some $N\in\mathbb{N}$ so that for all $n\geq N$, $|a_n^2-a^2|<\epsilon$. By algebra, $|a_n^2-a^2|=|a_n-a||a_n+a|$. Consider $|a_n+a|$. By the triangle inequality, $|a_n+a|\leq|a_n|+|a|$, thus $|a_n|$ is bounded by some $M\in\mathbb{N}$.

I know I'm trying to choose $M$ so that $|a_n-a|<\epsilon+M+|a|$. Where should I take it from here?

Answer 1

You're almost there, but the last sentence is in the wrong direction. Namely, you show that there is some $M\geq 0$ such that $$ |a_n+a| \leq |a|+M $$ for all $n$. For convenience, let $M' \stackrel{\rm def}{=} |a|+M$ (assume this is strictly positive, without loss of generality. We'll divide by $M'$... but if $M'=0$, then you're done as $a_n=a=0$ for all $n$ anyways). Now, this means that $$ |a_n^2-a^2| = |a_n-a||a_n+a| \leq |a_n-a|\cdot M' $$ This is multiplicative, not additive as you had in your last sentence. Now, all you need is to use the fact that $a_n\to 1$ to show that $|a_n-a| \leq \frac{\varepsilon}{M'}$ for $n$ large enough, since that'll imply $|a_n^2-a^2|\leq\frac{\varepsilon}{M'}\cdot M' = \varepsilon$.

Answer 2

For the case $a\neq 0$, assuming wlog $a>0$, we can also use that by definition of limit eventually

$$-2a\le a_n \le 2a \implies |a_n+a|\le 3 a$$

then

$$|a_n^2-a^2|=|a_n-a||a_n+a| \le 3a|a_n-a|$$

For the case $a=0$ we can use that eventually $|a_n|<1$.